Question

the two half - reactions are given below.
$2\mathrm{e}^- + 2\mathrm{h}_2\mathrm{o} + \mathrm{hpo}_3^{2 - } \
ightarrow \mathrm{h}_2\mathrm{po}_2^- + 3\mathrm{oh}^-$
$4\mathrm{oh}^- + \mathrm{n}_2\mathrm{h}_4 \
ightarrow \mathrm{n}_2 + 4\mathrm{h}_2\mathrm{o} + 4\mathrm{e}^-$
what is the coefficient for the hydroxide ion in the combined balanced net ionic equation?

Explanation:

Step1: Balance electrons in both half - reactions

The first half - reaction has 2 electrons gained: \(2e^-+2H_2O + HPO_3^{2 - }
ightarrow H_2PO_2^-+3OH^-\)
The second half - reaction has 4 electrons lost: \(4OH^-+N_2H_4
ightarrow N_2 + 4H_2O+4e^-\)
To balance the number of electrons, we multiply the first half - reaction by 2.
The first half - reaction after multiplication: \(4e^- + 4H_2O+2HPO_3^{2 - }
ightarrow 2H_2PO_2^-+6OH^-\)

Step2: Add the two half - reactions together

Now we add the modified first half - reaction and the second half - reaction:
\((4e^- + 4H_2O+2HPO_3^{2 - })+(4OH^-+N_2H_4)
ightarrow(2H_2PO_2^-+6OH^-)+(N_2 + 4H_2O+4e^-)\)
Cancel out the electrons (\(4e^-\) on both sides), cancel out \(4H_2O\) on both sides.
We get: \(2HPO_3^{2 - }+N_2H_4 + 4OH^-
ightarrow2H_2PO_2^-+N_2+6OH^-\)
Now, simplify the \(OH^-\) ions. Subtract \(4OH^-\) from both sides of the equation.
The net ionic equation: \(2HPO_3^{2 - }+N_2H_4
ightarrow2H_2PO_2^-+N_2 + 2OH^-\)

Answer:

The coefficient of the hydroxide ion (\(OH^-\)) in the balanced net ionic equation is 2.